In Tom Apostol Calculus: method of exhaustion for the area of a parabolic segment of $x^2$, proving $b^3/3$ is the only number between $s_n$ and $S_n$

Question

In the method of exhaustion for the area of a parabolic segment of $x^2$, the part where the author proves that $b^3/3$ is the only number which satisfies $s_n <A<S_n$

The prove has come down to this

$\frac{b^3}{3}-\frac{b^3}{n}< A<\frac{b^3}{3}+\frac{b^3}{n} $ for every integer $n>=1$. ,.....(1)

Now there are three possibilities

$ A>\frac{b^3}{3}$ , $ A<\frac{b^3}{3}$ , $ A<\frac{b^3}{3}$

The author now shows that the first two inequalities lead to a contradiction, then we must have $ A=\frac{b^3}{3}$.

(what I find difficult to understand is the contradiction, I can't see which fact is being contradicted.)

The author says this:

Suppose the inequality $ A>\frac{b^3}{3}$ were true. Then from the second inequality in (1) we obtain

$ A-\frac{b^3}{3}<\frac{b^3}{n}$ for every integer $n \geq 1$.

Since $ A-\frac{b^3}{3}$ is positive, we may divide both side by $ A-\frac{b^3}{3}$ and then multiply by n to obtain the equivalent statement $ n<\frac{b^3}{A-\frac{b^3}{3}}$ for every n.

But this inequality is obviously false when $ n \geq \frac{b^3}{A-\frac{b^3}{3}}$. Hence the inequality $ A>\frac{b^3}{3}$ leads to contradiction.

I am confused about what contradicts what?

Is it saying that $ A>\frac{b^3}{3}$ is true only when $ n<\frac{b^3}{A-\frac{b^3}{3}}$. and it contradicts from the assumption that it should be true for all $n$?

Answer 1

You have showed that, if we assume that $A > \frac{b^3}{3}$, then equation (1) is true only when $n < \frac{b^2}{A - \frac{b^3}{3}}$. However, the equation (1) must be true for all integers $n \geq 1$. We have reached a contradiction since our assumption implies an upper bound for $n$ in equation (1), but the equation must be true for all integer $n \geq 1$.